Question: Solve for $x$ : $ 3|x - 5| + 3 = -2|x - 5| + 4 $
Add $ {2|x - 5|} $ to both sides: $ \begin{eqnarray} 3|x - 5| + 3 &=& -2|x - 5| + 4 \\ \\ { + 2|x - 5|} && { + 2|x - 5|} \\ \\ 5|x - 5| + 3 &=& 4 \end{eqnarray} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} 5|x - 5| + 3 &=& 4 \\ \\ { - 3} &=& { - 3} \\ \\ 5|x - 5| &=& 1 \end{eqnarray} $ Divide both sides by ${5}$ $ \dfrac{5|x - 5|} {{5}} = \dfrac{1} {{5}} $ Simplify: $ |x - 5| = \dfrac{1}{5}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 5 = -\dfrac{1}{5} $ or $ x - 5 = \dfrac{1}{5} $ Solve for the solution where $x - 5$ is negative: $ x - 5 = -\dfrac{1}{5} $ Add ${5}$ to both sides: $ \begin{eqnarray} x - 5 &=& -\dfrac{1}{5} \\ \\ {+ 5} && {+ 5} \\ \\ x &=& -\dfrac{1}{5} + 5 \end{eqnarray} $ Change the ${ + 5}$ to an equivalent fraction with a denominator of $5$ $ x = - \dfrac{1}{5} {+ \dfrac{25}{5}} $ $ x = \dfrac{24}{5} $ Then calculate the solution where $x - 5$ is positive: $ x - 5 = \dfrac{1}{5} $ Add ${5}$ to both sides: $ \begin{eqnarray} x - 5 &=& \dfrac{1}{5} \\ \\ {+ 5} && {+ 5} \\ \\ x &=& \dfrac{1}{5} + 5 \end{eqnarray} $ Change the ${ + 5}$ to an equivalent fraction with a denominator of $5$ $ x = \dfrac{1}{5} {+ \dfrac{25}{5}} $ $ x = \dfrac{26}{5} $ Thus, the correct answer is $x = \dfrac{24}{5} $ or $x = \dfrac{26}{5} $.